Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $x = \dfrac{p + 9}{p + 4} \div \dfrac{-3p + 30}{p^2 - 6p - 40} $
Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{p + 9}{p + 4} \times \dfrac{p^2 - 6p - 40}{-3p + 30} $ First factor the quadratic. $x = \dfrac{p + 9}{p + 4} \times \dfrac{(p + 4)(p - 10)}{-3p + 30} $ Then factor out any other terms. $x = \dfrac{p + 9}{p + 4} \times \dfrac{(p + 4)(p - 10)}{-3(p - 10)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (p + 9) \times (p + 4)(p - 10) } { (p + 4) \times -3(p - 10) } $ $x = \dfrac{ (p + 9)(p + 4)(p - 10)}{ -3(p + 4)(p - 10)} $ Notice that $(p - 10)$ and $(p + 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ (p + 9)\cancel{(p + 4)}(p - 10)}{ -3\cancel{(p + 4)}(p - 10)} $ We are dividing by $p + 4$ , so $p + 4 \neq 0$ Therefore, $p \neq -4$ $x = \dfrac{ (p + 9)\cancel{(p + 4)}\cancel{(p - 10)}}{ -3\cancel{(p + 4)}\cancel{(p - 10)}} $ We are dividing by $p - 10$ , so $p - 10 \neq 0$ Therefore, $p \neq 10$ $x = \dfrac{p + 9}{-3} $ $x = \dfrac{-(p + 9)}{3} ; \space p \neq -4 ; \space p \neq 10 $